6.003 (Spring 2010)
April 7, 2010
Quiz #2
Name:
Kerberos Username:
Please circle your section number:
Section
1
2
3
4
Instructor
Time
Peter Hagelstein
Peter Hagelstein
Rahul Sarpeshkar
Rahul Sarpeshkar
10 am
11 am
1 pm
2 pm
Grades will be determined by the correctness of your answers (explanations
are not required).
Partial credit will be given for ANSWERS that demonstrate some but not
all of the important conceptual issues.
You have two hours.
Please put your initials on all subsequent sheets.
Enter your answers in the boxes.
This quiz is closed book, but you may use two 8.5 × 11 sheets of paper (four sides total).
No calculators, computers, cell phones, music players, or other aids.
1
/20
2
/30
3
/20
4
/30
Total
/100
Quiz #2 / 6.003: Signals and Systems (Spring 2010)
1. Convolution
2
[20 points]
Signals x1 (t) and x2 (t) are shown in the plots below, and are zero outside the indicated
intervals.
x1
1
t
0 1 2 3
5 6
8
10
12 13
15 16
18
x2
1
t
0 1 2 3
Plot the result of convolving x1 (t) with x2 (t).
Make sure that the important break-points are clear.
x1 ∗ x2
3
2
1
t
−5
5
−1
−2
−3
10
15
20
Quiz #2 / 6.003: Signals and Systems (Spring 2010)
4
[30 points]
2. Impulse response
Consider the following control system where the gain K is a real-valued constant.
Z t
+
X
−
Z t
dτ
K
dτ
−∞
Part a. [10 points] Determine the system function H(s) =
H(s) =
+
Y
−∞
s2
Y (s)
X(s)
as a function of K.
K(s + 1)
+ Ks + K
The forward signal path (from the output of the left adder to the output of the right adder) is
K 1s + K s12 = K s+1
s2 . By Black’s equation,
H(s) =
1
K(s+1)
s2
K(s+1)
+ s2
=
s2
K(s + 1)
+ Ks + K
Quiz #2 / 6.003: Signals and Systems (Spring 2010)
5
Part b. [20 points] The following plot shows the impulse response h(t) of the closed-loop
system for a particular value of K. [The dashed curves are exponential functions of time,
shown for reference.]
h(t)
4π seconds
t [seconds]
0
5
10
15
20
Determine K and indicate its value by placing an
25
30
on the following scale.
K
0.01
0.1
1
10
100
1000
The pole locations can also be found from the impulse response:
1
2π
1
1
1
s = ± jωd ≈ ± j
≈ ±j
τ
8
4π
8
2
where τ is the time constant of the exponential decay (approximately 8 seconds) and the frequency
ωd of the oscillation is 2π over the period of the oscillation (4π seconds). Then (s + 18 + j 12 )(s +
1
1
1
1
1
1
2
8 − j 2 ) = s + 4 s + 64 + 4 . Thus K ≈ 4 .
Quiz #2 / 6.003: Signals and Systems (Spring 2010)
6
[20 points]
3. Frequency Response
Let H(s) = Y (s)/X(s) for the following system.
X
+
−
K1
+
−
1
s − 30
K2
1
s+3
Y
Find K1 and K2 so that |H(jω)| matches the plot below.
|H(jω)| [log scale]
10
1
0.1
ω [log scale]
0.01
1
10
100
Enter numbers (or numerical expressions) for K1 and K2 in the boxes.
10
3
K1 =
K2 =
≈ 3.33
30
The system function for the inner loop is
1
K2
s−30
K2
+ s−30
=
K2
s + K2 − 30
Thus the overall system function is
(
(
1
K1 s+KK22−30
s+3
K 1 K2
(
(= 2
H(s) =
K2
1
s
+
(K
−
27)s
+ 3K2 − 90 + K1 K2
2
1 + K1 s+K2 −30
s+3
At low frequencies, the magnitude of the system function must be 1:
|H(j0)| =
K 1 K2
= 1.
3K2 − 90 + K1 K2
At ω = 100 (i.e., a high frequency), the magnitude of the system function must be 10−2 :
Quiz #2 / 6.003: Signals and Systems (Spring 2010)
|H(j100)| =
7
K 1 K2
K1 K2
=
= 10−2 .
ω2
1002
The latter implies that K1 K2 = 100, so that the former yields
and K1 = 100/30 = 10/3.
100
3K2 −90+100
= 1, so that K2 = 30
Quiz #2 / 6.003: Signals and Systems (Spring 2010)
9
[30 points]
4. Fourier series
Let x(t) represent a periodic signal (period T = 8 seconds) whose Fourier series coeffi­
cients are
ak =
1
jπk
0
for integers k = 0
k = 0.
When x(t) is the input to a linear, time-invariant system with system function H(s)
x(t)
H(s)
y(t)
the output y(t) is the sum of a constant β plus a periodic train of impulses with area 1,
y(t) = β +
∞
0
δ(t − 8k)
k=−∞
as shown below.
1
β
t
0
0
8
16
Part a. [15 points] Determine β.
−
β =
1
8
The Fourier series coefficients of y(t) are
bk =
1
8
(δ(t) + β)e−j
8
2π
8 kt
dt =
1
8
1
8
+β
k=0
k>0
However b0 must be zero, since a0 = 0. Therefore, β = − 81 .
24
Quiz #2 / 6.003: Signals and Systems (Spring 2010)
10
Part b. [15 points] Consider the output of the same system if the period of the in­
put signal is changed to T = 4 seconds while keeping the Fourier series coefficients ak
unchanged.
Is it possible to determine the new output signal from the information provided?
possible? (Yes or No)
Yes
From part a, we know the frequency response of the system at frequencies that are integer multiples
of 2π
8 radians/second:
k
ω
1
2π/8
bk
jπk
=
ak
8
jπ/8
2
4π/8
j2π/8
3
6π/8
j3π/8
4
8π/8
j4π/8
H(jω) =
...
For part b, we need to know the frequency response of the system at frequencies that are integer
multiples of 2π
4 radians/second, which is just every other line in the table above:
k
ω
H(jω)
1
2π/4
j2π/8
2
4π/4
j4π/8
3
6π/4
j6π/8
4
8π/4
j8π/8
...
Thus, each harmonic in the new signal is scaled by twice the previous scale factor.
The constant is therefore − 14 (twice − 18 ).
The areas of the impulses in part a were 1. Thus the magnitudes of the corresponding harmonics
were 18 . Doubling these magnitudes gives harmonics with magnitudes of 14 , so the areas of the
impulses are still 1.
If Yes, sketch and fully label the new output signal on the axes below.
1
0
− 14
t
4
8
12
16
20
24
Quiz #2 / 6.003: Signals and Systems (Spring 2010)
If No, briefly explain why not.
11
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6.003 Signals and Systems
Fall 2011
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